At the other end of what would come to be known as the Roman Empire, Parthian (Persian)

archers were able to discourage pursuers by firing backwards from the back of a galloping
horse-the famous "Parthian shot."
a. If the speed of the arrow vB the bowstring is vB, that of the horse is vH, the magnitude
of gravitational acceleration is g, and the angle above the horizontal at which the archer
aims is Delta B, what is the range of the shot, measured on the ground from the point of
the shot to the point the arrow falls? Disregard aerodynamic effects and the height
of the horse and rider. Assume the arrow leaves the bow at the angle Delta B at which it is
pointed, as seen by the archer.

b) For vB = 50:0 m/s, vH = 18:0 m/s, and g = 9:81 m/s^2, at what angle Delta B should the
archer aim to maximize the range of part a?

c) For the parameter values of part b, what is the maximum range of the shot, as described
in part a?
d. What is the maximum ground-to-ground range the archer could attain with the same
bow, standing on the ground? That is, how much range does the archer give up for
the Parthian shot? Again, ignore aerodynamic effects and the height of the archer.

To calculate the range of the Parthian shot, we can use the principles of projectile motion. Let's break down each sub-question and explain how to solve them step by step:

a) To find the range of the shot, we need to determine the horizontal distance traveled by the arrow before it falls to the ground. The key parameters needed are the initial velocity of the arrow (vB), the velocity of the horse (vH), the acceleration due to gravity (g), and the launch angle (Delta B).

b) To maximize the range, we need to determine the launch angle (Delta B) at which the horizontal distance is the greatest. Given the velocities vB, vH, and the acceleration due to gravity g, we can calculate the optimal angle for maximum range.

c) Once we have the launch angle (from part b), we can plug in the values of vB, vH, and g to find the maximum range of the shot.

d) Finally, we need to compare the range obtained from the Parthian shot with the maximum ground-to-ground range the archer could achieve by standing on the ground.

Now, let's solve each part in detail:

a) The range of the shot can be calculated using the formula:
Range = (vB^2 * sin(2*Delta B))/g

b) To maximize the range, we differentiate the range equation with respect to Delta B and set it equal to zero. This will give us the angle at which the range is maximum. We solve this equation to find Delta B.

c) Once the optimal angle (Delta B) is obtained from part b), we can substitute the values of vB, vH, and g into the range equation to calculate the maximum range.

d) To find the maximum ground-to-ground range the archer could attain while standing on the ground, we can use the range formula assuming a launch angle of 45 degrees (optimal for maximum range in projectile motion). Again, plug in the values of vB, g, and Delta B (45 degrees) into the range equation.

By comparing the results from parts c) and d), we can determine the range difference between the Parthian shot and standing on the ground.

I hope this explanation helps you understand how to solve each part of the question step by step.